Let $U,X,Y$ real random variable. If $U$ is independent of $X$ does $$\mathbb P(U\in A\mid X,Y)=\mathbb P(U\in A\mid X),$$ for all Borel set $A$ ? I think it's true, but a situation bother me. Take $(B_t)$ be a Brownian motion and $0<u<t$. So, $B_t-B_v$ is independent of $B_v$, but $B_t-B_v$ is obviously $\sigma (B_t,B_v)$ measurable. But writing $$\mathbb P(B_t-B_v\in A\mid B_t,B_v)=\mathbb P(B_t-B_v\in A\mid B_t)$$
looks strange. Indeed $\mathbb P(B_t-B_v\in A\mid B_t,B_v)=\boldsymbol 1_{\{B_t-B_v\in A\}}$ whereas $\mathbb P(B_t-B_v\in A\mid B_t)$ being $\sigma (B_t)-$measurable it doesn't look to be equal to $\boldsymbol 1_{\{B_t-B_v\in A\}}$. Can someone enlighten me ?